calculus - The Bi-linear Functional Equation $sum_{i=1}^{n}f_i(x)g_i(y)=0$

Consider the following so-called bi-linear functional equation
$$\sum_{i=1}^{n}f_i(x)g_i(y)=f_1(x)g_1(y)+f_2(x)g_2(y)+\cdot\cdot\cdot+f_n(x)g_n(y)=0 \tag{1}$$
where $f_i(x)$ and $g_i(y)$ are arbitrary functions.
What conditions are required for such a functional equation to hold?

For example, for $n=2$ we have

$$\begin{align} f_1(x)g_1(y)+f_2(x)g_2(y) &= 0 \\ \frac{f_1(x)}{f_2(x)}+\frac{g_2(y)}{g_1(y)} &= 0 \end{align} \tag{2}$$

where I assumed that $f_2(x) \ne 0$ and $g_1(y) \ne 0$. Consequently, we can conclude that Eq.$(2)$ will hold if and only if

$$\begin{align} f_1(x) &= +\lambda f_2(x) \\ g_2(y) &= -\lambda g_1(y) \end{align} \tag{3}$$

where $\lambda$ is some constant. Considering other cases, I ended up with

$$\boxed{ \begin{array}{ll} 1.\, \text{if} \, f_2(x)=0 \, \text{and} \, g_1(y)=0 \, \text{then} & 0=0 \\ 2.\, \text{if} \, f_2(x)\ne0 \, \text{and} \, g_1(y)=0 \, \text{then} & g_2(y)=0 \\ 3.\, \text{if} \, f_2(x)=0 \, \text{and} \, g_1(y)\ne0 \, \text{then} & f_1(x)=0 \\ 4.\, \text{if} \, f_2(x)\ne0 \, \text{and} \, g_1(y)\ne0 \, \text{then} & f_1(x) = +\lambda f_2(x) \, \text{and} \, g_2(y) = -\lambda g_1(y) \end{array} }\tag{4}$$

How can I generalize the result in $(4)$ for functional equation $(1)$?

Any help is appreciated.

Answer

I'll suppose $f_i$ and $g_i$ are defined on sets $D_f$ and $D_g$ respectively, taking values in a field $\mathbb F$ (you're probably thinking of $\mathbb F = \mathbb R$ or $\mathbb C$).

Suppose the linear span of $(f_1, \ldots, f_n)$ (as functions of $x \in D_f$) has dimension $r$. Thus the vectors $(c_1, \ldots, c_n)$ such that
$c_1 f_1 + \ldots + c_n f_n = 0$ form a subspace $V$ of $\mathbb F^n$ of dimension $n-r$. Then (1) is equivalent to: $(g_1(y), \ldots, g_n(y)) \in V$ for all $y \in D_g$.

For example, in the case $n=2$ your solution corresponds to the case $r=1$, where $V$ is spanned by $(1, -\lambda)$.